Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__F(g(X), Y) → A__F(mark(X), f(g(X), Y))
A__F(g(X), Y) → MARK(X)
MARK(g(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__F(x1, x2)  =  x1
g(x1)  =  g(x1)
mark(x1)  =  x1
f(x1, x2)  =  x1
MARK(x1)  =  x1
a__f(x1, x2)  =  x1

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(f(X1, X2)) → A__F(mark(X1), X2)
MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(f(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
f(x1, x2)  =  f(x1)
a__f(x1, x2)  =  a__f(x1)
g(x1)  =  g(x1)
mark(x1)  =  mark(x1)

Recursive Path Order [RPO].
Precedence:

[MARK₁, f₁, a_f1, g₁, mark₁]

The following usable rules [FROCOS05] were oriented:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__f(g(X), Y) → a__f(mark(X), f(g(X), Y))
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(g(X)) → g(mark(X))
a__f(X1, X2) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE